The rate constant is given by the equation \(= P.Ze^{–E/RT}\). Which factor should register a decrease for the reaction to proceed more rapidly?

E

The correct answer is option 3. \(E\).

In the equation for the rate constant, \(k = PZe^{(-E/RT)}\), a decrease in factor E, which represents the activation energy, will cause the reaction to proceed more rapidly.

The activation energy (E) is the minimum energy required for a chemical reaction to occur. It represents the energy barrier that reactant molecules must overcome in order to undergo the reaction and form products. By decreasing the activation energy, the energy barrier is lowered, allowing a larger fraction of reactant molecules to have sufficient energy to surpass the barrier and participate in the reaction.

A lower activation energy means that more reactant molecules will possess the required energy to react, resulting in a higher reaction rate. Thus, a decrease in the activation energy will make the reaction proceed more rapidly.

Therefore, the correct answer is (3) Ε, as a decrease in factor E (activation energy) will cause the reaction to proceed more rapidly.